Collection of basic formulas for a school chemistry course

Collection of basic formulas for a school chemistry course

G. P. Loginova

Elena Savinkina

E. V. Savinkina G. P. Loginova

Collection of basic formulas in chemistry

Student's Pocket Guide

general chemistry

The most important chemical concepts and laws

Chemical element- this is a certain type of atom with the same nuclear charge.

Relative atomic mass(A r) shows how many times the mass of an atom of a given chemical element is greater than the mass of a carbon-12 atom (12 C).

Chemical substance– a collection of any chemical particles.

Chemical particles

Formula unit– a conventional particle, the composition of which corresponds to the given chemical formula, for example:

Ar – argon substance (consists of Ar atoms),

H 2 O – the substance water (consists of H 2 O molecules),

KNO 3 – potassium nitrate substance (consists of K + cations and NO 3 ¯ anions).

Relationships between physical quantities

Atomic mass (relative) of the element B, A r (B):

Where *T(atom B) – mass of an atom of element B;

*t and– atomic mass unit;

*t and = 1/12 T(12 C atom) = 1.6610 24 g.

Quantity of substance B, n(B), mol:

Where N(B)– number of particles B;

N A– Avogadro's constant (N A = 6.0210 23 mol -1).

Molar mass of a substance V, M(V), g/mol:

Where t(V)– mass B.

Molar volume of gas IN, V M l/mol:

Where V M = 22.4 l/mol (a consequence of Avogadro’s law), under normal conditions (no. – atmospheric pressure p = 101,325 Pa (1 atm); thermodynamic temperature T = 273.15 K or Celsius temperature t = 0 °C).

B for hydrogen, D(gas B by H 2):

*Density of gaseous substance IN by air, D(gas B over air): Mass fraction of element E in matter V, w(E):

Where x is the number of E atoms in the formula of substance B

The structure of the atom and the Periodic Law D.I. Mendeleev

Mass number (A) – the total number of protons and neutrons in the atomic nucleus:

A = N(p 0) + N(p +).

Atomic nuclear charge (Z) equal to the number of protons in the nucleus and the number of electrons in the atom:

Z = N(p+) = N(e¯).

Isotopes– atoms of the same element, differing in the number of neutrons in the nucleus, for example: potassium-39: 39 K (19 p + , 20n 0, 19e¯); potassium-40: 40 K (19 p+, 21n 0, 19e¯).

*Energy levels and sublevels

*Atomic orbital(AO) characterizes the region of space in which the probability of an electron having a certain energy being located is greatest.

*Shapes of s- and p-orbitals

Periodic law and periodic system D.I. Mendeleev

The properties of elements and their compounds are periodically repeated with increasing atomic number, which is equal to the charge of the nucleus of the element’s atom.

Period number corresponds number of energy levels filled with electrons, and stands for the last energy level to be filled(EU).

Group number A shows And etc.

Group number B shows number of valence electrons ns And (n – 1)d.

S-elements section– the energy sublevel (ESL) is filled with electrons ns-EPU– IA- and IIA-groups, H and He.

p-elements section– filled with electrons np-EPU– IIIA-VIIIA-groups.

D-elements section– filled with electrons (P- 1) d-EPU – IB-VIIIB2-groups.

f-elements section– filled with electrons (P-2) f-EPU – lanthanides and actinides.

Changes in the composition and properties of hydrogen compounds of elements of the 3rd period of the Periodic Table

Non-volatile, decomposes with water: NaH, MgH 2, AlH 3.

Volatile: SiH 4, PH 3, H 2 S, HCl.

Changes in the composition and properties of higher oxides and hydroxides of elements of the 3rd period of the Periodic Table

Basic: Na 2 O – NaOH, MgO – Mg(OH) 2.

Amphoteric: Al 2 O 3 – Al(OH) 3.

Acidic: SiO 2 – H 4 SiO 4, P 2 O 5 – H 3 PO 4, SO 3 – H 2 SO 4, Cl 2 O 7 – HClO 4.

Chemical bond

Electronegativity(χ) is a quantity characterizing the ability of an atom in a molecule to acquire a negative charge.

Mechanisms of covalent bond formation

Exchange mechanism- the overlap of two orbitals of neighboring atoms, each of which had one electron.

Donor-acceptor mechanism– overlap of a free orbital of one atom with an orbital of another atom that contains a pair of electrons.

Overlap of orbitals during bond formation

*Type of hybridization – geometric shape of the particle – angle between bonds

Hybridization of central atom orbitals– alignment of their energy and form.

sp– linear – 180°

sp 2– triangular – 120°

sp 3– tetrahedral – 109.5°

sp 3 d– trigonal-bipyramidal – 90°; 120°

sp 3 d 2– octahedral – 90°

Mixtures and solutions

Solution- a homogeneous system consisting of two or more substances, the content of which can be varied within certain limits.

Solution: solvent (eg water) + solute.

True solutions contain particles smaller than 1 nanometer.

Colloidal solutions contain particles ranging in size from 1 to 100 nanometers.

Mechanical mixtures(suspensions) contain particles larger than 100 nanometers.

Suspension=> solid + liquid

Emulsion=> liquid + liquid

Foam, fog=> gas + liquid

Heterogeneous mixtures are separated settling and filtering.

Homogeneous mixtures are separated evaporation, distillation, chromatography.

Saturated solution is or may be in equilibrium with the solute (if the solute is solid, then its excess is in the precipitate).

Solubility– the content of the dissolved substance in a saturated solution at a given temperature.

Unsaturated solution less,

Supersaturated solution contains solute more, than its solubility at a given temperature.

Relationships between physicochemical quantities in solution

Mass fraction of solute IN, w(B); fraction of a unit or %:

Where t(V)– mass B,

t(r)– mass of solution.

Weight of solution, m(p), g:

m(p) = m(B) + m(H 2 O) = V(p) ρ(p),

where F(p) is the volume of the solution;

ρ(p) – solution density.

Volume of solution, V(p), l:

Molar concentration, s(V), mol/l:

Where n(B) is the amount of substance B;

M(B) – molar mass of substance B.

Changing the composition of the solution

Diluting the solution with water:

> t"(V)= t(B);

> the mass of the solution increases by the mass of added water: m"(p) = m(p) + m(H 2 O).

Evaporating water from a solution:

> the mass of the solute does not change: t"(B) = t(B).

> the mass of the solution decreases by the mass of evaporated water: m"(p) = m(p) – m(H 2 O).

Merging two solutions: The masses of solutions, as well as the masses of the dissolved substance, add up:

t"(B) = t(B) + t"(B);

t"(p) = t(p) + t"(p).

Crystal Drop: the mass of the solute and the mass of the solution are reduced by the mass of precipitated crystals:

m"(B) = m(B) – m(sediment); m"(p) = m(p) – m(sediment).

The mass of water does not change.

Thermal effect of a chemical reaction

*Enthalpy of formation of a substance ΔH°(B), kJ/mol, is the enthalpy of the reaction of formation of 1 mole of a substance from simple substances in their standard states, that is, at constant pressure (1 atm for each gas in the system or at a total pressure of 1 atm in the absence of gaseous reaction participants) and constant temperature (usually 298 K , or 25 °C).

*Thermal effect of a chemical reaction (Hess’s law)

Q = ΣQ(products) – ΣQ(reagents).

ΔН° = ΣΔН°(products) – Σ ΔН°(reagents).

For reaction aA + bB +… = dD + eE +…

ΔH° = (dΔH°(D) + eΔH°(E) +…) – (aΔH°(A) + bΔH°(B) +…),

Where a, b, d, e– stoichiometric amounts of substances corresponding to the coefficients in the reaction equation.

Chemical reaction rate

If during time τ in volume V the amount of reactant or product changed by Δ n, speed reaction:

For a monomolecular reaction A →…:

v = k c(A).

For the bimolecular reaction A + B → ...:

v = k c(A) c(B).

For the trimolecular reaction A + B + C → ...:

v = k c(A) c(B) c(C).

Changing the rate of a chemical reaction

Speed ​​reaction increase:

1) chemically active reagents;

2) promotion reagent concentrations;

3) increase

4) promotion temperature;

5) catalysts. Speed ​​reaction reduce:

1) chemically inactive reagents;

2) demotion reagent concentrations;

3) decrease surfaces of solid and liquid reagents;

4) demotion temperature;

5) inhibitors.

*Temperature speed coefficient(γ) is equal to a number that shows how many times the reaction rate increases when the temperature increases by ten degrees:

Chemical equilibrium

*Law of mass action for chemical equilibrium: in a state of equilibrium, the ratio of the product of the molar concentrations of products in powers equal to

Their stoichiometric coefficients, to the product of the molar concentrations of the reactants in powers equal to their stoichiometric coefficients, at a constant temperature is a constant value (concentration equilibrium constant).

In a state of chemical equilibrium for a reversible reaction:

aA + bB + … ↔ dD + fF + …

K c = [D] d [F] f .../ [A] a [B] b ...

*Shift in chemical equilibrium towards the formation of products

1) Increasing the concentration of reagents;

2) reducing the concentration of products;

3) increase in temperature (for an endothermic reaction);

4) decrease in temperature (for an exothermic reaction);

5) increase in pressure (for a reaction occurring with a decrease in volume);

6) decrease in pressure (for a reaction occurring with an increase in volume).

Exchange reactions in solution

Electrolytic dissociation– the process of formation of ions (cations and anions) when certain substances are dissolved in water.

acids are formed hydrogen cations And acid anions, For example:

HNO 3 = H + + NO 3 ¯

During electrolytic dissociation reasons are formed metal cations and hydroxide ions, for example:

NaOH = Na + + OH¯

During electrolytic dissociation salts(medium, double, mixed) are formed metal cations and acid anions, for example:

NaNO 3 = Na + + NO 3 ¯

KAl(SO 4) 2 = K + + Al 3+ + 2SO 4 2-

During electrolytic dissociation acid salts are formed metal cations and acid hydroanions, for example:

NaHCO 3 = Na + + HCO 3 ‾

Some strong acids

HBr, HCl, HClO 4, H 2 Cr 2 O 7, HI, HMnO 4, H 2 SO 4, H 2 SeO 4, HNO 3, H 2 CrO 4

Some strong reasons

RbOH, CsOH, KOH, NaOH, LiOH, Ba(OH) 2, Sr(OH) 2, Ca(OH) 2

Dissociation degree α– the ratio of the number of dissociated particles to the number of initial particles.

At constant volume:

Classification of substances by degree of dissociation

Berthollet's rule

Exchange reactions in solution proceed irreversibly if the result is the formation of a precipitate, gas, or weak electrolyte.

Examples of molecular and ionic reaction equations

1. Molecular equation: CuCl 2 + 2NaOH = Cu(OH) 2 ↓ + 2NaCl

“Complete” ionic equation: Сu 2+ + 2Сl¯ + 2Na + + 2OH¯ = Cu(OH) 2 ↓ + 2Na + + 2Сl¯

“Short” ionic equation: Cu 2+ + 2OH¯ = Cu(OH) 2 ↓

2. Molecular equation: FeS (T) + 2HCl = FeCl 2 + H 2 S

“Complete” ionic equation: FeS + 2H + + 2Сl¯ = Fe 2+ + 2Сl¯ + H 2 S

“Short” ionic equation: FeS (T) + 2H + = Fe 2+ + H 2 S

3. Molecular equation: 3HNO 3 + K 3 PO 4 = H 3 PO 4 + 3KNO 3

“Complete” ionic equation: 3H + + 3NO 3 ¯ + 3K + + PO 4 3- = H 3 PO 4 + 3K + + 3NO 3 ¯

“Short” ionic equation: 3H + + PO 4 3- = H 3 PO 4

*Hydrogen value

(pH) pH = – log = 14 + log

*pH range for dilute aqueous solutions

pH 7 (neutral environment)

Examples of exchange reactions

Neutralization reaction- an exchange reaction that occurs when an acid and a base interact.

1. Alkali + strong acid: Ba(OH) 2 + 2HCl = BaCl 2 + 2H 2 O

Ba 2+ + 2ON¯ + 2H + + 2Сl¯ = Ba 2+ + 2Сl¯ + 2Н 2 O

H + + OH¯ = H 2 O

2. Slightly soluble base + strong acid: Cu(OH) 2(t) + 2HCl = CuCl 2 + 2H 2 O

Cu(OH) 2 + 2H + + 2Cl¯ = Cu 2+ + 2Cl¯ + 2H 2 O

Cu(OH) 2 + 2H + = Cu 2+ + 2H 2 O

*Hydrolysis– an exchange reaction between a substance and water without changing the oxidation states of atoms.

1. Irreversible hydrolysis of binary compounds:

Mg 3 N 2 + 6H 2 O = 3Mg(OH) 2 + 2NH 3

2. Reversible hydrolysis of salts:

A) Salt is formed a strong base cation and a strong acid anion:

NaCl = Na + + Сl¯

Na + + H 2 O ≠ ;

Cl¯ + H 2 O ≠

There is no hydrolysis; neutral environment, pH = 7.

B) Salt is formed a strong base cation and a weak acid anion:

Na 2 S = 2Na + + S 2-

Na + + H 2 O ≠

S 2- + H 2 O ↔ HS¯ + OH¯

Hydrolysis by anion; alkaline environment, pH >7.

B) Salt is formed a cation of a weak or slightly soluble base and an anion of a strong acid:

End of introductory fragment.

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Magnitude and its dimension	Ratio
Atomic mass of element X (relative)
Element serial number	Z= N(e –) = N(R +)
Mass fraction of element E in substance X, in fractions of a unit, in %)
Amount of substance X, mol
Amount of gas substance, mol	V m= 22.4 l/mol (n.s.) Well. – R= 101 325 Pa, T= 273 K
Molar mass of substance X, g/mol, kg/mol
Mass of substance X, g, kg	m(X) = n(X) M(X)
Molar volume of gas, l/mol, m 3 /mol	V m= 22.4 l/mol at N.S.
Gas volume, m3	V = V m × n
Product yield
Density of substance X, g/l, g/ml, kg/m3
Density of gaseous substance X by hydrogen
Density of gaseous substance X in air	M(air) = 29 g/mol
United Gas Law
Mendeleev-Clapeyron equation	PV = nRT, R= 8.314 J/mol×K
Volume fraction of a gaseous substance in a mixture of gases, in fractions of a unit or in %
Molar mass of a mixture of gases
Mole fraction of a substance (X) in a mixture
Amount of heat, J, kJ	Q = n(X) Q(X)
Thermal effect of reaction	Q =–H
Heat of formation of substance X, J/mol, kJ/mol
Chemical reaction rate (mol/lsec)
Law of Mass Action (for a simple reaction)	a A+ V B= With C + d D u = k With a(A) With V(B)
Van't Hoff's rule
Solubility of the substance (X) (g/100 g solvent)
Mass fraction of substance X in mixture A + X, in fractions of a unit, in %
Weight of solution, g, kg	m(rr) = m(X)+ m(H2O) m(rr) = V(rr)  (rr)
Mass fraction of dissolved substance in solution, in fractions of a unit, in %
Solution density
Volume of solution, cm 3, l, m 3
Molar concentration, mol/l
Degree of electrolyte dissociation (X), in fractions of a unit or %
Ionic product of water	K(H2O) =
pH value	pH = –lg

Main:

Kuznetsova N.E. and etc. Chemistry. 8th grade-10th grade. – M.: Ventana-Graf, 2005-2007.

Kuznetsova N.E., Litvinova T.N., Levkin A.N. Chemistry.11th grade in 2 parts, 2005-2007.

Egorov A.S. Chemistry. A new textbook for preparing for higher education. Rostov n/d: Phoenix, 2004.– 640 p.

Egorov A.S. Chemistry: a modern course for preparing for the Unified State Exam. Rostov n/a: Phoenix, 2011. (2012) – 699 p.

Egorov A.S. Self-instruction manual for solving chemical problems. – Rostov-on-Don: Phoenix, 2000. – 352 p.

Chemistry/tutor manual for applicants to universities. Rostov-n/D, Phoenix, 2005– 536 p.

Khomchenko G.P., Khomchenko I.G.. Problems in chemistry for applicants to universities. M.: Higher school. 2007.–302p.

Additional:

Vrublevsky A.I.. Educational and training materials for preparing for centralized testing in chemistry / A.I. Vrublevsky –Mn.: Unipress LLC, 2004. – 368 p.

Vrublevsky A.I.. 1000 problems in chemistry with chains of transformations and control tests for schoolchildren and applicants. – Mn.: Unipress LLC, 2003. – 400 p.

Egorov A.S.. All types of calculation problems in chemistry for preparation for the Unified State Exam. – Rostov n/D: Phoenix, 2003. – 320 p.

Egorov A.S., Aminova G.Kh.. Typical tasks and exercises for preparing for the chemistry exam. – Rostov n/d: Phoenix, 2005. – 448 p.

Unified State Exam 2007. Chemistry. Educational and training materials for preparing students / FIPI - M.: Intellect-Center, 2007. – 272 p.

Unified State Exam 2011. Chemistry. Educational and training kit ed. A.A. Kaverina. – M.: National Education, 2011.

The only real options for tasks to prepare for the Unified State Exam. Unified State Examination 2007. Chemistry/V.Yu. Mishina, E.N. Strelnikova. M.: Federal Testing Center, 2007.–151 p.

Kaverina A.A. The optimal bank of tasks for preparing students. Unified State Exam 2012. Chemistry. Textbook./ A.A. Kaverina, D.Yu. Dobrotin, Yu.N. Medvedev, M.G. Snastina. – M.: Intellect-Center, 2012. – 256 p.

Litvinova T.N., Vyskubova N.K., Azhipa L.T., Solovyova M.V.. Test tasks in addition to tests for students of 10-month correspondence preparatory courses (methodological instructions). Krasnodar, 2004. – P. 18 – 70.

Litvinova T.N.. Chemistry. Unified State Exam 2011. Training tests. Rostov n/d: Phoenix, 2011.– 349 p.

Litvinova T.N.. Chemistry. Tests for the Unified State Exam. Rostov n/d.: Phoenix, 2012. - 284 p.

Litvinova T.N.. Chemistry. Laws, properties of elements and their compounds. Rostov n/d.: Phoenix, 2012. - 156 p.

Litvinova T.N., Melnikova E.D., Solovyova M.V.., Azhipa L.T., Vyskubova N.K. Chemistry in tasks for applicants to universities. – M.: Onyx Publishing House LLC: Mir and Education Publishing House LLC, 2009. – 832 p.

Educational and methodological complex in chemistry for students of medical and biological classes, ed. T.N. Litvinova. – Krasnodar.: KSMU, – 2008.

Chemistry. Unified State Exam 2008. Entrance tests, teaching aid / ed. V.N. Doronkina. – Rostov n/a: Legion, 2008.– 271 p.

List of websites on chemistry:

1. Alhimik. http:// www. alhimik. ru

2. Chemistry for everyone. Electronic reference book for a complete chemistry course.

http:// www. informika. ru/ text/ database/ chemistry/ START. html

3. School chemistry - reference book. http:// www. schoolchemistry. by. ru

4. Chemistry tutor. http://www. chemistry.nm.ru

Internet resources

Alhimik. http:// www. alhimik. ru

Chemistry for everyone. Electronic reference book for a complete chemistry course.

http:// www. informika. ru/ text/ database/ chemistry/ START. html

School chemistry - reference book. http:// www. schoolchemistry. by. ru

http://www.classchem.narod.ru

Chemistry tutor. http://www. chemistry.nm.ru

http://www.alleng.ru/edu/chem.htm- educational Internet resources on chemistry

http://schoolchemistry.by.ru/- school chemistry. This site has the opportunity to take On-line testing on various topics, as well as demo versions of the Unified State Exam

Chemistry and life—XXI century: popular science magazine. http:// www. hij. ru

Modern symbols for chemical elements were introduced into science in 1813 by J. Berzelius. According to his proposal, elements are designated by the initial letters of their Latin names. For example, oxygen (Oxygenium) is designated by the letter O, sulfur (Sulfur) by the letter S, hydrogen (Hydrogenium) by the letter H. In cases where the names of the elements begin with the same letter, one more letter is added to the first letter. Thus, carbon (Carboneum) has the symbol C, calcium (Calcium) - Ca, copper (Cuprum) - Cu.

Chemical symbols are not only abbreviated names of elements: they also express certain quantities (or masses), i.e. Each symbol represents either one atom of an element, or one mole of its atoms, or a mass of an element equal to (or proportional to) the molar mass of that element. For example, C means either one carbon atom, or one mole of carbon atoms, or 12 mass units (usually 12 g) of carbon.

Chemical formulas

Formulas of substances also indicate not only the composition of the substance, but also its quantity and mass. Each formula represents either one molecule of a substance, or one mole of a substance, or a mass of a substance equal to (or proportional to) its molar mass. For example, H2O represents either one molecule of water, or one mole of water, or 18 mass units (usually (18 g) of water.

Simple substances are also designated by formulas showing how many atoms a molecule of a simple substance consists of: for example, the formula for hydrogen H 2. If the atomic composition of a molecule of a simple substance is not precisely known or the substance consists of molecules containing a different number of atoms, and also if it has an atomic or metallic structure rather than a molecular one, the simple substance is designated by the symbol of the element. For example, the simple substance phosphorus is denoted by the formula P, since, depending on conditions, phosphorus can consist of molecules with a different number of atoms or have a polymer structure.

Chemistry formulas for solving problems

The formula of the substance is determined based on the results of the analysis. For example, according to analysis, glucose contains 40% (wt.) carbon, 6.72% (wt.) hydrogen and 53.28% (wt.) oxygen. Therefore, the masses of carbon, hydrogen and oxygen are in the ratio 40:6.72:53.28. Let us denote the desired formula for glucose C x H y O z, where x, y and z are the numbers of carbon, hydrogen and oxygen atoms in the molecule. The masses of the atoms of these elements are respectively equal to 12.01; 1.01 and 16.00 amu Therefore, the glucose molecule contains 12.01x amu. carbon, 1.01u amu hydrogen and 16.00zа.u.m. oxygen. The ratio of these masses is 12.01x: 1.01y: 16.00z. But we have already found this relationship based on glucose analysis data. Hence:

12.01x: 1.01y: 16.00z = 40:6.72:53.28.

According to the properties of proportion:

x: y: z = 40/12.01:6.72/1.01:53.28/16.00

or x:y:z = 3.33:6.65:3.33 = 1:2:1.

Therefore, in a glucose molecule there are two hydrogen atoms and one oxygen atom per carbon atom. This condition is satisfied by the formulas CH 2 O, C 2 H 4 O 2, C 3 H 6 O 3, etc. The first of these formulas - CH 2 O- is called the simplest or empirical formula; it has a molecular weight of 30.02. In order to find out the true or molecular formula, it is necessary to know the molecular mass of a given substance. When heated, glucose is destroyed without turning into gas. But its molecular weight can be determined by other methods: it is equal to 180. From a comparison of this molecular weight with the molecular weight corresponding to the simplest formula, it is clear that the formula C 6 H 12 O 6 corresponds to glucose.

Thus, a chemical formula is an image of the composition of a substance using symbols of chemical elements, numerical indices and some other signs. The following types of formulas are distinguished:

— simplest , which is obtained experimentally by determining the ratio of chemical elements in a molecule and using the values ​​of their relative atomic masses (see example above);

— molecular , which can be obtained by knowing the simplest formula of a substance and its molecular weight (see example above);

— rational , displaying groups of atoms characteristic of classes of chemical elements (R-OH - alcohols, R - COOH - carboxylic acids, R - NH 2 - primary amines, etc.);

— structural (graphic) , showing the relative arrangement of atoms in a molecule (can be two-dimensional (in a plane) or three-dimensional (in space));

— electronic, displaying the distribution of electrons across orbitals (written only for chemical elements, not for molecules).

Let's take a closer look at the example of the ethyl alcohol molecule:

1. the simplest formula of ethanol is C 2 H 6 O;
2. the molecular formula of ethanol is C 2 H 6 O;
3. the rational formula of ethanol is C 2 H 5 OH;

Examples of problem solving

EXAMPLE 1

Exercise	With complete combustion of an oxygen-containing organic substance weighing 13.8 g, 26.4 g of carbon dioxide and 16.2 g of water were obtained. Find the molecular formula of a substance if the relative density of its vapors with respect to hydrogen is 23.
Solution	Let’s draw up a diagram of the combustion reaction of an organic compound, designating the number of carbon, hydrogen and oxygen atoms as “x”, “y” and “z”, respectively: C x H y O z + O z →CO 2 + H 2 O. Let us determine the masses of the elements that make up this substance. Values ​​of relative atomic masses taken from the Periodic Table of D.I. Mendeleev, round to whole numbers: Ar(C) = 12 amu, Ar(H) = 1 amu, Ar(O) = 16 amu. m(C) = n(C)×M(C) = n(CO 2)×M(C) = ×M(C); m(H) = n(H)×M(H) = 2×n(H 2 O)×M(H) = ×M(H); Let's calculate the molar masses of carbon dioxide and water. As is known, the molar mass of a molecule is equal to the sum of the relative atomic masses of the atoms that make up the molecule (M = Mr): M(CO 2) = Ar(C) + 2×Ar(O) = 12+ 2×16 = 12 + 32 = 44 g/mol; M(H 2 O) = 2×Ar(H) + Ar(O) = 2×1+ 16 = 2 + 16 = 18 g/mol. m(C) = ×12 = 7.2 g; m(H) = 2 × 16.2 / 18 × 1 = 1.8 g. m(O) = m(C x H y O z) - m(C) - m(H) = 13.8 - 7.2 - 1.8 = 4.8 g. Let's determine the chemical formula of the compound: x:y:z = m(C)/Ar(C) : m(H)/Ar(H) : m(O)/Ar(O); x:y:z = 7.2/12:1.8/1:4.8/16; x:y:z = 0.6: 1.8: 0.3 = 2: 6: 1. This means the simplest formula of the compound is C 2 H 6 O and the molar mass is 46 g/mol. The molar mass of an organic substance can be determined using its hydrogen density: M substance = M(H 2) × D(H 2) ; M substance = 2 × 23 = 46 g/mol. M substance / M(C 2 H 6 O) = 46 / 46 = 1. This means the formula of the organic compound will be C 2 H 6 O.
Answer	C2H6O

EXAMPLE 2

Exercise	The mass fraction of phosphorus in one of its oxides is 56.4%. The oxide vapor density in air is 7.59. Determine the molecular formula of the oxide.
Solution	The mass fraction of element X in a molecule of the composition NX is calculated using the following formula: ω (X) = n × Ar (X) / M (HX) × 100%. Let's calculate the mass fraction of oxygen in the compound: ω(O) = 100% - ω(P) = 100% - 56.4% = 43.6%. Let us denote the number of moles of elements included in the compound as “x” (phosphorus), “y” (oxygen). Then, the molar ratio will look like this (the values ​​of relative atomic masses taken from D.I. Mendeleev’s Periodic Table are rounded to whole numbers): x:y = ω(P)/Ar(P) : ω(O)/Ar(O); x:y = 56.4/31: 43.6/16; x:y = 1.82:2.725 = 1:1.5 = 2:3. This means that the simplest formula for combining phosphorus with oxygen will be P 2 O 3 and a molar mass of 94 g/mol. The molar mass of an organic substance can be determined using its air density: M substance = M air × D air; M substance = 29 × 7.59 = 220 g/mol. To find the true formula of an organic compound, we find the ratio of the resulting molar masses: M substance / M(P 2 O 3) = 220 / 94 = 2. This means that the indices of phosphorus and oxygen atoms should be 2 times higher, i.e. the formula of the substance will be P 4 O 6.
Answer	P4O6

several basic concepts and formulas.

All substances have different mass, density and volume. A piece of metal from one element can weigh many times more than an exactly the same size piece of another metal.

Mole(number of moles)

designation: mole, international: mol- a unit of measurement for the amount of a substance. Corresponds to the amount of substance that contains N.A. particles (molecules, atoms, ions) Therefore, a universal quantity was introduced - number of moles. A frequently encountered phrase in tasks is “received... mole of substance"

N.A.= 6.02 1023

N.A.- Avogadro's number. Also “a number by agreement.” How many atoms are there in the tip of a pencil? About a thousand. It is not convenient to operate with such quantities. Therefore, chemists and physicists around the world agreed - let’s designate 6.02 × 1023 particles (atoms, molecules, ions) as 1 mole substances.

1 mole = 6.02 1023 particles

This was the first of the basic formulas for solving problems.

Molar mass of a substance

Molar mass substance is the mass of one mole of substance.

Denoted as Mr. It is found according to the periodic table - it is simply the sum of the atomic masses of a substance.

For example, we are given sulfuric acid - H2SO4. Let's calculate the molar mass of a substance: atomic mass H = 1, S-32, O-16.
Mr(H2SO4)=1 2+32+16 4=98 g\mol.

The second necessary formula for solving problems is

substance mass formula:

That is, to find the mass of a substance, you need to know the number of moles (n), and we find the molar mass from the Periodic Table.

Law of conservation of mass - The mass of substances that enter into a chemical reaction is always equal to the mass of the resulting substances.

If we know the mass(es) of the substances that reacted, we can find the mass(es) of the products of that reaction. And vice versa.

The third formula for solving chemistry problems is

volume of substance:

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Where did the number 22.4 come from? From Avogadro's law:

equal volumes of different gases taken at the same temperature and pressure contain the same number of molecules.

According to Avogadro's law, 1 mole of an ideal gas under normal conditions (n.s.) has the same volume Vm= 22.413 996(39) l

That is, if in the problem we are given normal conditions, then, knowing the number of moles (n), we can find the volume of the substance.

So, basic formulas for solving problems in chemistry

Avogadro's numberN.A.

6.02 1023 particles

Quantity of substance n (mol)

n=V\22.4 (l\mol)

Mass of substance m (g)

Volume of substance V(l)

V=n 22.4 (l\mol)

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These are formulas. Often, to solve problems, you first need to write the reaction equation and (required!) arrange the coefficients - their ratio determines the ratio of moles in the process.

Key words: Chemistry 8th grade. All formulas and definitions, symbols of physical quantities, units of measurement, prefixes for designating units of measurement, relationships between units, chemical formulas, basic definitions, briefly, tables, diagrams.

1. Symbols, names and units of measurement
some physical quantities used in chemistry

Physical quantity	Designation	Unit
Time	t	With
Pressure	p	Pa, kPa
Quantity of substance	ν	mole
Mass of substance	m	kg, g
Mass fraction	ω	Dimensionless
Molar mass	M	kg/mol, g/mol
Molar volume	Vn	m 3 /mol, l/mol
Volume of substance	V	m 3, l
Volume fraction		Dimensionless
Relative atomic mass	A r	Dimensionless
	M r	Dimensionless
Relative density of gas A to gas B	D B (A)	Dimensionless
Density of matter	R	kg/m 3, g/cm 3, g/ml
Avogadro's constant	N A	1/mol
Absolute temperature	T	K (Kelvin)
Temperature in Celsius	t	°C (degrees Celsius)
Thermal effect of a chemical reaction	Q	kJ/mol

2. Relationships between units of physical quantities

3. Chemical formulas in 8th grade

4. Basic definitions in 8th grade

• Atom- the smallest chemically indivisible particle of a substance.
• Chemical element- a certain type of atom.
• Molecule- the smallest particle of a substance that retains its composition and chemical properties and consists of atoms.
• Simple substances- substances whose molecules consist of atoms of the same type.
• Complex substances- substances whose molecules consist of atoms of different types.
• Qualitative composition of the substance shows which atoms of elements it consists of.
• Quantitative composition of the substance shows the number of atoms of each element in its composition.
• Chemical formula- conventional recording of the qualitative and quantitative composition of a substance using chemical symbols and indices.
• Atomic mass unit(amu) - a unit of measurement of atomic mass, equal to the mass of 1/12 of a carbon atom 12 C.
• Mole- the amount of a substance that contains a number of particles equal to the number of atoms in 0.012 kg of carbon 12 C.
• Avogadro's constant (Na = 6*10 23 mol -1) - the number of particles contained in one mole.
• Molar mass of a substance (M ) is the mass of a substance taken in an amount of 1 mole.
• Relative atomic mass element A r - the ratio of the mass of an atom of a given element m 0 to 1/12 of the mass of a carbon atom 12 C.
• Relative molecular weight substances M r - the ratio of the mass of a molecule of a given substance to 1/12 of the mass of a carbon atom 12 C. The relative molecular mass is equal to the sum of the relative atomic masses of the chemical elements forming the compound, taking into account the number of atoms of a given element.
• Mass fraction chemical element ω(X) shows what part of the relative molecular mass of substance X is accounted for by a given element.

ATOMIC-MOLECULAR TEACHING
1. There are substances with molecular and non-molecular structure.
2. There are gaps between the molecules, the sizes of which depend on the state of aggregation of the substance and temperature.
3. Molecules are in continuous motion.
4. Molecules are made up of atoms.
6. Atoms are characterized by a certain mass and size.
During physical phenomena, molecules are preserved; during chemical phenomena, as a rule, they are destroyed. Atoms rearrange during chemical phenomena, forming molecules of new substances.

LAW OF CONSTANT COMPOSITION OF MATTER
Each chemically pure substance of molecular structure, regardless of the method of preparation, has a constant qualitative and quantitative composition.

VALENCE
Valence is the property of an atom of a chemical element to attach or replace a certain number of atoms of another element.

CHEMICAL REACTION
A chemical reaction is a phenomenon as a result of which other substances are formed from one substance. Reactants are substances that enter into a chemical reaction. Reaction products are substances formed as a result of a reaction.
Signs of chemical reactions:
1. Release of heat (light).
2. Change in color.
3. Odor appears.
4. Formation of sediment.
5. Gas release.